Complex - Testing Convergence for $\sum_{n=1}^{\infty} n^{2}z^{n}$

Just wanting to make sure my conclusions are correct for this

$$\sum_{n=1}^{\infty} n^{2}z^{n}\quad \text{for}\quad |z| < 1$$

When using the ratio and root test, i conclude that my series has either divergence or convergence. Totally not cool right?

So I then and use Dirichlet Test https://en.wikipedia.org/wiki/Dirichlet%27s_test

Letting $a_n = n^{2}$ and $b_n = z^{2}$, its obvious that the function fails the first two tests.

So will it then diverge?

• If $|z|<1$, both the Ratio Test and the Root Test conclusively show that the series converges. – alex.jordan Mar 26 '16 at 19:43
• We like to help but there is some turbulence on this website! Whenever I post they just downvote which is an immature act! – Mhenni Benghorbal Mar 26 '16 at 19:45
• And actually, for $\sum_{n\geq0} P(n)z^n$, where $P$ is a polynomial or a rational fraction (with nonvanishing denominator for integer $n$), the radius of convergence is always $1$, by the above test. – Jean-Claude Arbaut Mar 26 '16 at 19:46
• @MhenniBenghorbal : Change your name on the website. Nothing is disturbed by doing that, except that someone targeting you may not know who you are any more. :) – DisintegratingByParts Mar 26 '16 at 21:59
• @TrialAndError One has to change the account then. Changing the printed name won't help, since it's possible to find the account with only the user number (try this), and it would be also possible to check an old question with an answer from the same person, where the name would be apparent. I fear the only solution is to start from scratch, which is rather annoying. – Jean-Claude Arbaut Mar 26 '16 at 22:30

The ratio test: $$1 > \lim_{n}\frac{(n+1)^2}{n^2}|z| =|z| .$$ The root test: $$1 > \lim_{n}(n^2|z|^{n})^{1/n} = \lim_{n}e^{2\frac{ln(n)}{n}}|z|=|z|.$$ In either case, you have absolute convergence for $|z| < 1$ and divergence for $|z| > 1$.